A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the [110] direction. If the critical resolved shear stress for this material is 1.75 MPa, calculate the magnitude(s) of applied stresses necessary to cause slip to occur on the (111) plane in each of the [11 0], [101] and [011] directions.

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.                                          

Explanation:        

Given:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

To calculate:

The magnitude of applied stress in the direction of [101] and [011].

Formula:

zcr=σ cosФ cosλ

Solution:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ  = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.                                                                                                                                      

the clapeyron equation (also called the clausius-clapeyron equation) relates the slope of a reaction line on a phase diagram to fundamental thermodynamic properties.

explanation: